/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 Consider an irreducible finite M... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider an irreducible finite Markov chain with states \(0,1, \ldots, N\). (a) Starting in state \(i\), what is the probability the process will ever visit state \(j\) ? Explain! (b) Let \(x_{i}=P\) [visit state \(N\) before state 0 |start in \(\left.i\right\\}\). Compute a set of linear equations that the \(x_{i}\) satisfy, \(i=0,1, \ldots, N\). (c) If \(\sum_{j} j P_{i j}=i\) for \(i=1, \ldots, N-1\), show that \(x_{i}=i / N\) is a solution to the equations in part (b).

Short Answer

Expert verified
In summary, to answer the question: (a) The probability of ever visiting state \(j\) from state \(i\) is given by the sum: \(\sum_{n = 1}^{\infty}f_{ij}^{(n)}\). (b) The set of linear equations for \(x_i\), the probability of visiting state \(N\) before state \(0\), is given by: \(x_{i} = \sum_{j = 1}^{N-1} P_{i j} x_j\). (c) If \(\sum_{j} j P_{i j}=i\) for \(i=1, \ldots, N-1\), then \(x_i = i/N\) is a solution to the equations in part (b).

Step by step solution

01

Define First Passage Probability

First passage probability is the probability that a system will reach a given state for the first time after a certain number of steps. Let \(f_{ij}^{(n)}\) denote the probability that the process will visit state \(j\) for the first time in \(n\) steps, starting from state \(i\).
02

(a) Probability of ever visiting state \(j\) from state \(i\)

To find the probability of ever visiting state \(j\) from state \(i\), we can sum up the probabilities of visiting state \(j\) for the first time at any number of steps. That is, $$\sum_{n = 1}^{\infty}f_{ij}^{(n)}$$ The probability of ever visiting state \(j\) from state \(i\) can be found using this sum.
03

(b) Find linear equations for \(x_i\)

Let \(x_i\) be the probability of visiting state \(N\) before state \(0\), starting from state \(i\). Notice that \(x_0 = 0\) (since we are already in state 0) and \(x_N = 1\) (since we are already in state \(N\)). Now, let's set up a system of linear equations based on the Markov chain properties. For each state \(i\), we can write: $$x_{i} = \sum_{j = 1}^{N-1} P_{i j} x_j$$ Here, \(P_{ij}\) is the transition probability from state \(i\) to state \(j\). Since \(i\) can range from \(1\) to \(N-1\), we can construct a set of linear equations with \(N - 1\) unknowns.
04

(c) Prove \(x_i = i/N\) is a solution

We are given that \(\sum_{j} j P_{i j}=i\) for \(i=1, \ldots, N-1\). Our goal is to show that if this condition holds, then the previously derived set of linear equations has the solution \(x_i = i / N\). For each state \(i\), we have: $$x_{i} = \sum_{j = 1}^{N-1} P_{i j} x_j = \sum_{j = 1}^{N-1} P_{i j} \frac{j}{N}$$ Now, replace the summation with the given condition: $$x_i = \frac{1}{N} \sum_{j} j P_{i j} = \frac{i}{N}$$ This shows that \(x_i = i / N\) is indeed a solution to the set of linear equations derived in part (b), given the specified condition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Passage Probability
In a Markov chain, one important concept is the first passage probability, which helps us understand how likely it is for a system to reach a specific state for the first time over a number of steps. Consider starting from state \(i\) and wanting to reach state \(j\). The probability that this will occur for the first time at step \(n\) is denoted as \(f_{ij}^{(n)}\). This concept is crucial for predicting system behavior over time.

This first occurrence can be calculated by evaluating an infinite sum of probabilities:
  • \(\sum_{n=1}^{\infty} f_{ij}^{(n)}\)
This sum gives the overall probability that you'll eventually reach state \(j\) if you started from state \(i\). Transitioning in a probabilistic setting like a Markov chain means that direct prediction is challenging, but first passage probability allows for measuring reachability over time.
Transition Probability
Transition probability is a key component of Markov chains and refers to the likelihood of moving from one state to another. In the context of a Markov chain, this is denoted as \(P_{ij}\), which represents the probability of transitioning from state \(i\) to state \(j\) in one step.

The matrix formed by transition probabilities (often called the transition matrix) contains all these individual probabilities. In a system with states \(0, 1, \ldots, N\):
  • The element at row \(i\) and column \(j\) of this matrix is \(P_{ij}\).
  • Each row sums up to 1, showing that the total probability distribution over possible transitions from a given state is complete.
Understanding these probabilities is essential because they lay the groundwork for analyzing the flow and dynamics within the Markov process. They help in calculating important metrics such as the first passage probabilities and in setting up the linear equations needed to solve specific probabilistic predictions.
Linear Equations
Linear equations in the context of Markov chains are used to find specific probabilities, such as the likelihood of reaching a certain state before another. For instance, let \(x_i\) represent the probability of reaching state \(N\) before state \(0\) when starting from state \(i\). Setting up these equations involves using the transition probabilities \(P_{ij}\).

Consider:
  • \(x_0 = 0\) because starting at state \(0\) means it's impossible to reach state \(0\) again "before" it's reached.
  • \(x_N = 1\) since you're already at state \(N\).
  • For each intermediate state \(i\), the equation is \(x_{i} = \sum_{j=1}^{N-1} P_{ij} x_j\).
This set of equations allows us to solve for the unknown probabilities \(x_i\). Solving these kinds of linear equations helps in predicting system behavior, such as shown in the exercise where it is demonstrated that \(x_i = i/N\) can be a solution given certain conditions. The use of linear equations simplifies handling complex probabilistic systems by utilizing linear algebra techniques.

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Most popular questions from this chapter

On a chessboard compute the expected number of plays it takes a knight, starting in one of the four corners of the chessboard, to return to its initial position if we assume that at each play it is equally likely to choose any of its legal moves. (No other pieces are on the board.) Hint: Make use of Example 4.36.

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